Question

$\int (\sqrt[3]{x})\left(\sqrt[5]{1+\sqrt[3]{x^4}}\right)dx$ is equal to

• A. ${\left(1+x^{\frac{3}{4}}\right)}^{\frac{5}{6}}+c$
• B. ${\left(1+x^{\frac{4}{3}}\right)}^{\frac{5}{6}}+c$
• C. $\frac{5}{8}{\left(1+x^{\frac{4}{3}}\right)}^{\frac{6}{5}}+c$
• D. $\frac{1}{6}{\left(1+x^{\frac{4}{3}}\right)}^6+c$
• E. $\frac{15}{8}{\left(1+x^{\frac{4}{3}}\right)}^{\frac{6}{5}}+c$

Answer 1

Answer: (C)

Let $I=\int \sqrt[3]{x}(\sqrt[5]{1+\sqrt[3]{x^4}})dx$
Put $\sqrt[3]{x^4}=t$
$\Rightarrow$ $\frac{4}{3}.\sqrt[3]{x}dx=dt$
$\therefore$ $I=\frac{3}{4}\int (\sqrt[5]{1+t})dt$
$=\frac{3}{4}\left[\frac{{(1+t)}^{\frac{1}{5}+1}}{\frac{1}{5}+1}\right]+c$
$=\frac{5}{8}[{(1+\sqrt[3]{x^4})}^{6/5}]+c$